Question: You have found the following ages (in years) of all 5 tigers at your local zoo: $ 5,\enspace 1,\enspace 17,\enspace 1,\enspace 15$ What is the average age of the tigers at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 5 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{5 + 1 + 17 + 1 + 15}{{5}} = {7.8\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $5$ years $-2.8$ years $7.84$ years $^2$ $1$ year $-6.8$ years $46.24$ years $^2$ $17$ years $9.2$ years $84.64$ years $^2$ $1$ year $-6.8$ years $46.24$ years $^2$ $15$ years $7.2$ years $51.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{7.84} + {46.24} + {84.64} + {46.24} + {51.84}} {{5}} $ $ {\sigma^2} = \dfrac{{236.8}}{{5}} = {47.36\text{ years}^2} $ The average tiger at the zoo is 7.8 years old. The population variance is 47.36 years $^2$.